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[Max White]

1. 501
2. 418.
3. Open
4. Well, 969 are closed, do I really need to find the open ones?

This was done with some really lazy maths, not bothering to put my formulae together, and the fact that I had to to math means I almost certainly missed the point, sorry but I think I failed this one.

[Darvi]

Actually, multiples of perfect squares are also open. Perfect cubes and their multiples are closed. And so on.

[Urist Imiknorris]

Actually, multiples of perfect squares are also open.

This doesn't hold for all of them (and I offer 1 as a counterexample)

brb, writing a program to check.

[Max White]

Man, I looked at my 'Vector' card and I thought to myself, na, I'm building a Joker deck, when am I ever going to need a Vector card? Now I'm paying the price.

[Urist Imiknorris]

Spoiler (click to show/hide)

Those are the ones that will be open after the 1000^th student.

EDIT: Added total.

[Max White]

That is totally cheating. 

Then again, I remember once using a program to prove that you should always switch doors when shown a goat.

[Urist Imiknorris]

(I was just proving Darvi wrong)

[ed boy]

I've got one more of these questions.

This question concerns calendar dates of the form
d1d2/m1m2/y1y2y3y4
in the order day/month/year.
The question specifically concerns those dates which contain no repetitions of a digit.
For example, the date 23/05/1967 is one such date but 07/12/1974 is not such a date as both 1 = m1 = y1 and 7 = d2 = y3 are repeated digits.
We will use the Gregorian Calendar throughout (this is the calendar system that is standard throughout most of the world; see below.)
(i) Show that there is no date with no repetition of digits in the years from 2000 to 2099.
(ii) What was the last date before today with no repetition of digits?
(iii) When will the next such date be?
(iv) How many such dates were there in years from 1900 to 1999?
[The Gregorian Calendar uses 12 months, which have, respectively, 31, 28 or 29, 31, 30,
31, 30, 31, 31, 30, 31, 30 and 31 days. The second month (February) has 28 days in
years that are not divisible by 4, or that are divisible by 100 but not 400 (such as 1900);
it has 29 days in the other years (leap years).]

[Darvi]

Well huh.

[Urist Imiknorris]

(i) Show that there is no date with no repetition of digits in the years from 2000 to 2099.

The month will contain a zero (which is repeated in the year) in all cases except for November and December. In November, 1 repeats itself in the month, and in December, the 2 in the month is repeated in the year:

??/01/20??
??/02/20??
??/03/20??
??/04/20??
??/05/20??
??/06/20??
??/07/20??
??/08/20??
??/09/20??
??/10/20??
??/11/20??
??/12/20??

(ii) What was the last date before today with no repetition of digits?

25/06/1987 (June 25, 1987)

(iii) When will the next such date be?

17/06/2345 (June 17, 2345)

(iv) How many such dates were there in years from 1900 to 1999?

360.

[Kogut]

<post to watch/>

[ed boy]

You have a bunch of regular six-sided dice, with each dice being fair.

In each round, you roll your dice at once. Each time one of them comes up a 1, it is removed from your collection of dice and it is not rolled again in subsequent rounds. If it does not come up a 1, it is rolled again in the next round.

If you start the first round with N dice, how many rounds would you expect to go through before you eliminate all your dice?

[Darvi]

Infinite.

[ed boy]

Infinite.

Nope

[Darvi]

I always expect the worst case scenario :V